a) Write the balanced chemical equation for the complete combustion of 1 mole of

Question

a) Write the balanced chemical equation for the complete combustion of 1 mole of liquid methanol (CH3OH) to carbon dioxide (CO2 (g)) and water (H2O(g)).

(b) Using the data given, calculate the standard heat of reaction (combustion) when methanol is completely burned in air.

Data: Delta H0f H2O(gas) = -242.1KJ/mol, delta H of CO2(gas) = -39308KJ/mol, delta H of CH3OH(l) = - 238.7kj/mol

(c) using the answer obtained in (b) and the data given below, calculate the standard heat of reaction (combustion) of methonal at 300 degree C.

Data: Mean specific heat capacities over the range 25 degree C to 300 degree C for CO2 = 32.0 J mol-1K-1 , H2O(g) 36.5J mol-1 K-1, O2 = 30.0J/mol/K, CH3OH(l) = 50 J/mol/K

(d) (i) using the answer from (b) calculate a value for the heat of reaction of methanol at constant volume, Take R = 0.0083KJ/mol/K

(ii) give one instance where the heat of reaction at constant volume my be required rather than the one at constant pressure.

Explanation / Answer

a]

2CH3OH + 3O2 ------> 4H2O + 2CO2

b]

delta Hrxn = delta H of products - delta H of reactants

delta Hrxn = [4*delta H of H2O + 2*delta H of CO2] - [3*delta H of O2 + 2*delta H of CH3OH)

delta Hrxn = -1277.16 KJ

c]

delta H rxn at 300 Kelvin is

delta Hrxn (300) = delta Hrxn(298K) + [ Cp*dT]300298   

Substitute the values .....U will get the answer [ Cp values are given ]

d]

At constant VOlume

delta Hrxn = Change in internal energy [ delta E]

delta H = delta E + delta n*RT

delta n = No of products - no of reactants

delta n = 6-5 = 1

We already given R value and T is 298 K

delta E = -1277.16 + 0.0083*298 = -1274.6866 KJ

So enthalpy is -1274.6866 KJ

b]

During sublimation

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